The normal airflow over the Rocky Mountains is west to east. The air loses much
ID: 1702255 • Letter: T
Question
The normal airflow over the Rocky Mountains is west to east. The air loses much of its moisture content and is chilled as it climbs the western side of the mountains. When it descends on the eastern side, the increase in pressure toward lower altitudes causes the temperature to increase. The flow, then called a chinook wind, can rapidly raise the air temperature at the base of the mountains. Assume that the air pressure p depends on altitude y according to p = p0exp(-ay), where p0 = 1.00 atm and a = 1.15 10-4 m-1. Also assume that the ratio of the molar specific heats is ? = 4/3. A parcel of air with an initial temperature of -5.00°C descends adiabatically from y1 = 4057 m to y = 1567 m. What is its temperature at the end of the descent?Explanation / Answer
Initial pressure Po = 1 atm
air pressure at the altitude y is P = Po e (-ay)
a = 1.15*10-4 m-1
ratio of molar specific heats r = 4 / 3
initial temperature of the air T1 = - 5 o C
Altitude y1 = 4057 m
y = 1567 m
pressure at altitude P1 = Po e-ay1
= ( 1 atm ) ( e -(1.15*10^ -4 )(4057m) = 0.628 atm
pressure at altitude at y P2 = Po e-ay
=( 1 atm ) ( e -(1.15*10^ -4 )(1567m) = 0.835 atm
initial temperature Ti = - 5 o C
= 268 k
in adiabatic process the relation ship between pressure and temperature is P1-r Tr = k
( T2 / T1 ) r= ( P2 / p1 ) r-1
T2 = T1 ( P2 / p1 ) 1-r
= (268)(0.7520) 1-4/3
= 284.721
= 21.71 degrees
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